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Torsion and the conservation of angular momentum

  1. May 16, 2010 #1
    If our spacetime has torsion, will it violate the conservation of angular momentum?

    Thx
     
  2. jcsd
  3. May 16, 2010 #2
    It's difficult to tell what you mean. Are you asking what would happen if empty space had torsion, or if we used a connection that was not torsion free? --or if general relativity properly handles orbital angular momentum?
     
    Last edited: May 16, 2010
  4. May 16, 2010 #3
    I guess you mean the torsion tensor [tex]S^{i}{}_{k l} \equiv \Gamma^{i}{}_{k l} - \Gamma^{i}{}_{l k}[/itex]. According to the principle of Relativity, it must be zero.
     
  5. May 17, 2010 #4
    I'm unclear about this, Dickfore. In the derivation I'm familiar with, the torsion tensor must be zero. How does one even begin to set up a connection that is not symmetric on interchange of the lower indices?
     
  6. May 17, 2010 #5
    In GR, the spacetime is Riemann manifold, with zero torsion. But it is just a assumption. If there is not some experimental fact conflict with the torsion , we still cannot rule out the spacetime with torsion which is called Riemann Cartan manifold.

    People introduce this assumption for what? Just for simplification? (forgive my poor English,:) )
     
  7. May 18, 2010 #6
    I agree; I thought the "zero torsion" was just an assumption. So I'd like to here more about this statement:
    Why would non-zero torsion violate the principle of relativity? Are you saying measurements validating the principle of relativity can, in principle, rule out Einstein-Cartan gravity?
     
  8. May 18, 2010 #7
    micomaco86572

    By coincidence, I've been searching the internet for Einstein Cartan Gravity and related structures the past week.

    -------------------------------------------------------------------------------------------------

    To define the covariant derivative, a connection, [itex]\Gamma[/itex] is needed.

    [tex]\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma_{\mu \lambda}^{\nu} V^{\lambda}[/tex]

    [itex]\nabla_{\mu}[/itex] acts linearly on vectors and [itex]\nabla_\mu V^\nu[/itex] transforms as the product of a vector and dual vector.

    [tex]\nabla_{\mu'} V^{\nu'} = \frac{ \partial x^{\mu} } { \partial x^{\mu'} } \frac{ \partial x^{\nu'} }{ \partial x^{\nu} } \nabla_{\mu} V^{\nu}[/tex]

    In examining the transformation of [itex]\Gamma[/itex], it is apparent that it is symmetric in its lower indices. I won't post the transformation equation that shows this because it's too much work to get all this Latex correct, though it's apparent that it is unchanged upon interchange of the lower indices.

    It is symmetrical in lower indices,

    [tex]\Gamma_{\mu \lambda}^{\nu} = \Gamma_{\lambda \mu}^{\nu} \ ,[/tex]

    so the difference, S must be zero in all its parts.

    [tex]S_{\mu \lambda}^{\nu} = \Gamma_{\mu \lambda}^{\nu} - \Gamma_{\lambda \mu}^{\nu} = 0[/tex]


    So how does one obtain a nonzero torsion, S?

    If a nonzero torsion is ad hoc--where an antisymmetric connection is simply added onto the symmetric one, how can the covariant derivative be justified if it is only defined upon the symmetric part?
     
    Last edited: May 18, 2010
  9. May 18, 2010 #8
    It seems to violate the Equivalence Principle. An accelerated particle will not begin to spin. If I understand correctly, Einstein Cartan gravity will allow this. The verbage I read was hazzy and no mathematics was given.
     
    Last edited: May 18, 2010
  10. May 18, 2010 #9
    According to http://en.wikipedia.org/wiki/Einstein%E2%80%93Cartan_theory" [Broken],

    "general relativity has one known flaw: it cannot describe "spin-orbit coupling", i.e., exchange of intrinsic angular momentum (spin) and orbital angular momentum...".
     
    Last edited by a moderator: May 4, 2017
  11. May 19, 2010 #10
    I am sorry, Phrak. I cannot prove the symmetry of the lower indices. Can u tell me some details. Thx a lot.
     
  12. May 19, 2010 #11
    I can direct you to it instead. Go to this web site http://xxx.lanl.gov/abs/gr-qc/9712019" The download is 1.51 MB. See equation 3.6.

    I'm sorry, I've made a grave error. Thank you for bringing it to my attention.

    Equation 3.6 says the connection Coordinate Transformation is symmetric in its lower indices.

    Equation 3.6 does Not say the connection is symmetric.

    If you absorb equations 3.1 through 3.6, continue on to equation 3.16 that defines the torsion tensor.
     
    Last edited by a moderator: Apr 25, 2017
  13. May 19, 2010 #12
    micomaco. I found one justification for using a torsion-free connection in Sean Carroll's Text on page 120:

    "We could drop the demand that the connection be torsion-free, in which case the torsion tensor could lead to additional propagating degrees of freedom. Without going into details, the basic reason why such theories do not receive much attention is simply because the torsion is itself a tensor; there is nothing to distinguish it from other, “non-gravitational” tensor fields. Thus, we do not really lose any generality by considering theories of torsion-free connections (which lead to GR) plus any number of tensor fields, which we can name what we like."
     
  14. May 21, 2010 #13
    micomaco86572, have you found the answer to your question?

    I've done some review. There are a number of restrictions to obtain the covariant derivative for General Relativity (and therefore the connection as well). Restricting the Christoffel connection to torsion free is one of many restrictions.

    1) Linear operator: [tex] \nabla (S+T) = \nabla S + \nabla T[/tex]

    2) Product rule of differentiation: [tex]\nabla (S \otimes T) = (\nabla S) \otimes T + S \otimes (\nabla T)[/tex]

    3) Metric compatable (the metric commutes with the covariant derivative): [tex]\nabla_{\lambda} g_{_\mu \nu} = 0[/tex]

    4) Torsion free (the connection is symmetric in its lower indices): [tex]\Gamma_{\mu \nu}^{\lambda} = \Gamma_{\nu \mu}^{\lambda}[/tex]

    5) Acting on scalars, the covariant derivative reduces to the partial derivative: [tex]\nabla_{\mu} \phi = \partial{_\mu} \phi[/tex]

    6) And finally, nabla commutes with contraction: [tex]\nabla_{\mu} T_{\nu}\ ^{\nu} = (\nabla T)_{\mu \nu}\ ^{\nu}[/tex]

    I think all of these restrictions are to ensure the Equivalence Principle, including torsion=0, are requirements so that the geodesic, when defined as the parallel transport of a vector in its own direction, is the path of a freely falling body in an accelerating reference frame, and that the orientation of the body is also preserved as experiment indicates.
     
    Last edited: May 21, 2010
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